[LeetCode] Verify Preorder Serialization of a Binary Tree

最新推荐文章于 2024-08-26 07:28:39 发布

转载最新推荐文章于 2024-08-26 07:28:39 发布 · 63 阅读

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原文链接：http://www.cnblogs.com/immjc/p/9132692.html

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#数据结构与算法

本文介绍了一种不重建树即可验证给定字符串是否为有效二叉树前序遍历序列的方法。通过计数节点差的方式，判断字符串9,3,4,#,#,1,#,#,2,#,6,#,#等是否符合二叉树序列化规范。

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #.

     _9_
    /   \
   3     2
  / \   / \
 4   1  #  6
/ \ / \   / \
# # # #   # #

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where #represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character '#' representing nullpointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".

Example 1:

Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: "1,#"
Output: false

Example 3:

Input: "9,#,#,1"
Output: false

根据二叉树性质，空结点比实结点大一。

class Solution {
public:
    bool isValidSerialization(string preorder) {
        stringstream ss(preorder);
        string val;
        int diff = 1;
        while (getline(ss, val, ','))
        {
            if (diff == 0)
                return false;
            if (val == "#")
                diff--;
            else
                diff++;
        }
        return diff == 0;
    }
};